Returning address of a local variable directly using address of operator vs using pointer variable

Question

I'm aware of the fact that you should never return the address of a local variable from a function. But while demonstrating the fact I faced a problem. Consider the following program:

int *test()
{
    int x = 12;
    int *p = &x;
    return p;
}

int main()
{
    int *p = test();
    printf("%p",p);
    return 0;
}

It prints an address like 0061fed0 as expected. But if I return the address of x directly using & i.e. if the test function is changed as follows:

int *test()
{
    int x = 12;
    return &x;
}

then the output becomes 00000000. So, can you please explain what is happening here? I'm using gcc compiler that comes bundled with Code:: Blocks 17.12 IDE in windows 10.

Regarding Duplicate Questions: Suggested duplicate question: C - GCC generates wrong instructions when returning local stack address Explains the behaviour of using the address of operator directly but doesn't address the scenario where a pointer variable is used to return the address of a local variable, which is explained here in StoryTeller's answer: "The value of a pointer becomes indeterminate when the object it points to (or just past) reaches the end of its lifetime".

Possible duplicate of [C - GCC generates wrong instructions when returning local stack address](https://stackoverflow.com/questions/36967840/c-gcc-generates-wrong-instructions-when-returning-local-stack-address) — Sander De Dycker, Sep 11 '19 at 06:33
also : [Why does gcc return 0 instead of the address of a stack allocated variable?](https://stackoverflow.com/questions/40290049/why-does-gcc-return-0-instead-of-the-address-of-a-stack-allocated-variable) — Sander De Dycker, Sep 11 '19 at 06:36
Undefined behavior is behavior which is not defined. The compiler isn't obliged to produce any meaningful code. Speculating about why it generated a certain flavour of meaningless code isn't particularly interesting. — Lundin, Sep 11 '19 at 06:56
@Lundin "The value of a pointer becomes indeterminate when the object it points to (or just past) reaches the end of its lifetime." I didn't know that. That's why I asked the question. — Imran Rana, Sep 11 '19 at 11:35

score 4 · Accepted Answer · answered Sep 11 '19 at 06:30

Strictly speaking, from the point of view of C language specification, that's a valid outcome.

6.2.4 Storage durations of objects (emphasis mine)

2 The lifetime of an object is the portion of program execution during which storage is guaranteed to be reserved for it. An object exists, has a constant address, and retains its last-stored value throughout its lifetime. If an object is referred to outside of its lifetime, the behavior is undefined. The value of a pointer becomes indeterminate when the object it points to (or just past) reaches the end of its lifetime.

Therefore the value that function returns is indeterminate in either case. You cannot predict what it will be, or even use it in a meaningful way. A program that uses an indeterminate value has undefined behavior, anything can occur.

So what your compiler does is return null when you return the address of a local directly. It's a valid value according to the C language itself, since the object is dead soon anyway. But it has the benefit of likely crashing your program early in a modern hosted implementation, and allowing you to fix a bug.

"The value of a pointer becomes indeterminate when the object it points to (or just past) reaches the end of its lifetime." - That's what I was looking for. Thanks. — Imran Rana, Sep 11 '19 at 11:32

Returning address of a local variable directly using address of operator vs using pointer variable

1 Answers1